Question

A baby's mouth is 30 cm from her father's ear and 1.50 m from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Answer 1

Answer:

The difference between the sound intensity is 13.98 dB.

Explanation:

Given that,

Distance from father = 30 cm

Distance from mother = 1.50 cm

We need to calculate the intensity

Using formula of intensity

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

For father,

$I_{f}= \dfrac{P}{4\pi\times(30\times10^{-2})^2}$....(I)

For mother,

$I_{m}= \dfrac{P}{4\pi\times(1.50)^2}$....(II)

The difference between the sound intensity

$I'=10 log(\dfrac{I}{I_{o}})$

$I'=10 log(\dfrac{I_{f}}{I_{m}})$

Put the value into the formula

$I'=10 log(\dfrac{(1.50)^2}{(30\times10^{-2})^2})$

$I'=13.98\ dB$

Hence, The difference between the sound intensity is 13.98 dB.